Difference between revisions of "2014 AMC 12B Problems/Problem 25"
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<math> 2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1</math> | <math> 2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1</math> | ||
− | <math> \textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D) | + | <math> \textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi </math> |
− | ==Solution== | + | ==Solution 1== |
− | Rewrite <math>\cos{4x} - 1</math> as <math>2\cos^2{2x} - 2</math>. Now let <math>a = \cos{2x}</math>, and let <math>b = \cos{\left( \frac{2014\pi^2}{x} \right) }</math>. We have | + | Rewrite <math>\cos{4x} - 1</math> as <math>2\cos^2{2x} - 2</math>. Now let <math>a = \cos{2x}</math>, and let <math>b = \cos{\left( \frac{2014\pi^2}{x} \right) }</math>. We have: |
<cmath>2a(a - b) = 2a^2 - 2</cmath> | <cmath>2a(a - b) = 2a^2 - 2</cmath> | ||
− | <cmath>ab = 1</cmath> | + | |
+ | Therefore, <cmath>ab = 1</cmath>. | ||
+ | |||
Notice that either <math>a = 1</math> and <math>b = 1</math> or <math>a = -1</math> and <math>b = -1</math>. For the first case, <math>a = 1</math> only when <math>x = k\pi</math> and <math>k</math> is an integer. <math>b = 1</math> when <math>\frac{2014\pi^2}{k\pi}</math> is an even multiple of <math>\pi</math>, and since <math>2014 = 2*19*53</math>, <math>b =1</math> only when <math>k</math> is an odd divisor of <math>2014</math>. This gives us these possible values for <math>x</math>: | Notice that either <math>a = 1</math> and <math>b = 1</math> or <math>a = -1</math> and <math>b = -1</math>. For the first case, <math>a = 1</math> only when <math>x = k\pi</math> and <math>k</math> is an integer. <math>b = 1</math> when <math>\frac{2014\pi^2}{k\pi}</math> is an even multiple of <math>\pi</math>, and since <math>2014 = 2*19*53</math>, <math>b =1</math> only when <math>k</math> is an odd divisor of <math>2014</math>. This gives us these possible values for <math>x</math>: | ||
<cmath>x= \pi, 19\pi, 53\pi, 1007\pi</cmath> | <cmath>x= \pi, 19\pi, 53\pi, 1007\pi</cmath> | ||
For the case where <math>a = -1</math>, <math>\cos{2x} = -1</math>, so <math>x = \frac{m\pi}{2}</math>, where m is odd. <math>\frac{2014\pi^2}{\frac{m\pi}{2}}</math> must also be an odd multiple of <math>\pi</math> in order for <math>b</math> to equal <math>-1</math>, so <math>\frac{4028}{m}</math> must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for <math>m</math>, and therefore no cases where <math>a = -1</math> and <math>b = -1</math>. Therefore, the sum of all our possible values for <math>x</math> is | For the case where <math>a = -1</math>, <math>\cos{2x} = -1</math>, so <math>x = \frac{m\pi}{2}</math>, where m is odd. <math>\frac{2014\pi^2}{\frac{m\pi}{2}}</math> must also be an odd multiple of <math>\pi</math> in order for <math>b</math> to equal <math>-1</math>, so <math>\frac{4028}{m}</math> must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for <math>m</math>, and therefore no cases where <math>a = -1</math> and <math>b = -1</math>. Therefore, the sum of all our possible values for <math>x</math> is | ||
<cmath>\pi + 19\pi + 53\pi + 1007\pi = \boxed{\textbf{(D)}\ 1080 \pi}</cmath> | <cmath>\pi + 19\pi + 53\pi + 1007\pi = \boxed{\textbf{(D)}\ 1080 \pi}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Very similar to the solution above, re-write the expression using <math>\cos 4x = 2 \cos^2 2x - 1</math>: | ||
+ | |||
+ | <cmath>2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = 2 \cos^2 2x - 2</cmath> | ||
+ | |||
+ | Now, expand the LHS and cancel terms: | ||
+ | |||
+ | <cmath>2 \cos^2 2x - 2 \cos 2x \cos{\left( \frac{2014\pi^2}{x} \right) } = 2 \cos^2 2x - 2</cmath> | ||
+ | |||
+ | <cmath> \cos 2x \cos{\left( \frac{2014\pi^2}{x} \right) } = 1</cmath> | ||
+ | |||
+ | Now we use product-to-sum identities to get: | ||
+ | |||
+ | <cmath>\frac{1}{2} \left(\cos{2x + \left( \frac{2014\pi^2}{x} \right) } + \cos{2x + \left( \frac{2014\pi^2}{x} \right) } \right) = 1</cmath> | ||
+ | |||
+ | <cmath>\cos{2x + \left( \frac{2014\pi^2}{x} \right) } + \cos{2x + \left( \frac{2014\pi^2}{x} \right) } = 1</cmath> | ||
+ | |||
+ | Notice that for any <math>\theta</math>, <math>\max{\cos \theta} = 1</math>. This is achieved when <math>\theta = 0</math>, or equivalently | ||
+ | |||
+ | <cmath>2x - \frac{2014\pi^2}{x} \equiv 0 \mod 2\pi</cmath> | ||
+ | |||
+ | We can cleverly assume <math>x=c\pi</math> for some real <math>c</math>. Then, we must have | ||
+ | |||
+ | <cmath>2c - \frac{2014}{c} \equiv 0 \mod 2</cmath> | ||
+ | |||
+ | In order for this to be satisfied, <math>\frac{2014}{c}</math> must be an even integer. Factoring <math>2014 = 2 \cdot 19 \cdot 53</math>, we see that <math>c</math> must be a positive odd integer that divides <math>2014</math>. Our only positive valid <math>c</math> are <math>c = 1, 19, 53, 1007</math>. Our answer is just <math>\pi(1+19+53+1007) = 1080\pi \implies \textbf{(D)}</math>. | ||
+ | |||
+ | |||
+ | -FIREDRAGONMATH16 | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Rewriting <math>\cos{4x} - 1</math> as <math>2\cos^2{2x} - 2</math> and transposing <math>2\cos{2x}</math> from the LHS to the RHS, we get, | ||
+ | |||
+ | <cmath>\cos{2x} - \cos{\left(\frac{2014\pi^2}{x}\right)} = 1 - \frac{1}{\cos{2x}}</cmath> | ||
+ | <cmath>\implies \underbrace{\cos{2x} + \frac{1}{\cos{2x}}}_{\text{LHS}} = \underbrace{1 + \cos{\left(\frac{2014\pi^2}{x}\right)}}_{\text{RHS}}</cmath> | ||
+ | |||
+ | By the [https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality AM-GM Inequality], | ||
+ | |||
+ | <cmath>\cos{2x} + \frac{1}{\cos{2x}} \in (-\infty, -2] \cup [2, \infty)</cmath> | ||
+ | |||
+ | Also, because of the range of <math>\cos</math>, | ||
+ | |||
+ | <cmath>1 + \cos{\left(\frac{2014\pi^2}{x}\right)} \in [0, 2]</cmath> | ||
+ | |||
+ | Hence, <math>\text{LHS} = \text{RHS} = 2</math>, and we get (<math>m, n \in \mathbb{Z}</math>), | ||
+ | |||
+ | * <math>\cos{2x} = 1 \implies x = m\pi \hspace{75pt} \cdots\cdots (1)</math> | ||
+ | * <math>\cos{\left(\frac{2014\pi^2}{x}\right)} = 1 \implies x = \frac{1007\pi}{n} \hspace{22pt} \cdots\cdots (2)</math> | ||
+ | |||
+ | |||
+ | From <math>(1)</math> and <math>(2)</math>, | ||
+ | <cmath>m = \frac{1007}{n}</cmath> | ||
+ | <cmath>\implies m \in \{1, 19, 53, 1007\}</cmath> | ||
+ | <cmath>\implies x \in \{1\pi, 19\pi, 53\pi, 1007\pi\}</cmath> | ||
+ | |||
+ | Therefore, sum of values of <math>x</math> is <cmath>\pi + 19\pi + 53\pi + 1007\pi = \boxed{\textbf{(D)}\ 1080\pi}</cmath>. | ||
+ | |||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Plusone plusone] | ||
== See also == | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=24|after=Last Question}} | {{AMC12 box|year=2014|ab=B|num-b=24|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 09:08, 17 August 2023
Problem
Find the sum of all the positive solutions of
Solution 1
Rewrite as . Now let , and let . We have:
Therefore, .
Notice that either and or and . For the first case, only when and is an integer. when is an even multiple of , and since , only when is an odd divisor of . This gives us these possible values for : For the case where , , so , where m is odd. must also be an odd multiple of in order for to equal , so must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for , and therefore no cases where and . Therefore, the sum of all our possible values for is
Solution 2
Very similar to the solution above, re-write the expression using :
Now, expand the LHS and cancel terms:
Now we use product-to-sum identities to get:
Notice that for any , . This is achieved when , or equivalently
We can cleverly assume for some real . Then, we must have
In order for this to be satisfied, must be an even integer. Factoring , we see that must be a positive odd integer that divides . Our only positive valid are . Our answer is just .
-FIREDRAGONMATH16
Solution 3
Rewriting as and transposing from the LHS to the RHS, we get,
By the AM-GM Inequality,
Also, because of the range of ,
Hence, , and we get (),
From and ,
Therefore, sum of values of is .
~ plusone
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
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